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This is the second of our many quizzes on Galvanic cells. Click on this to access more such competitive quizzes on Electrochemistry.

Galvanic cells Quiz 2

chemistry

 Which of the following reactions reduces all metal in a given reaction? $ \\ M{{g}^{2+}}+2{{e}^{-}}\to Mg\left( s \right),\,\,\,\,\,E_{cell}^{{}^\circ }=-2.37V,$ $N{{i}^{2+}}+2{{e}^{-}}\to Ni\left( s \right), \,\,\,\,\,E_{cell}^{{}^\circ }=-0.26V,$ $ \\ Li\to L{{i}^{+}}+{{e}^{-}}=E_{cell}^{{}^\circ }=+3.04V$

 Lithium metal has the positive   $E_{cell}^{{}^\circ }$  or maximum value of   $E_{cell}^{{}^\circ }$ as compare to other metals value and hence can reduce all the other ions.

 To minimise all of the less negative values, use a high negative value. The reduction potential of lithium is $-3.04$. It will lower the levels of $M{{g}^{2+}}$ and $N{{i}^{2+}}$ ions. $ \\ \text{2Li}\left( s \right)+M{{g}^{2+}}\left( aq. \right)\to 2L{{i}^{+}}\left( aq. \right)+Mg\left( s \right) \\ \text{2Li}\left( s \right)+N{{i}^{2+}}\left( aq. \right)\to 2L{{i}^{+}}\left( aq. \right)+Ni\left( s \right) $

 Which of the following equations about the Nernst equation for a reaction is correct? $ \\ S{{n}^{2+}}\left( aq \right)+C{{u}^{2+}}\left( aq \right)\to Cu\left( s \right)+S{{n}^{4+}}\left( aq \right)$

  Using Nernst equation $ \\ {{E}_{cell}}=E_{cell}^{{}^\circ }-\dfrac{2.303RT}{nF}\log \dfrac{\left[ {{P}^{n+}} \right]}{\left[ {{R}^{n+}} \right]} \\ $ Where ${{P}^{n+}}\to $ product ion concentration ${{R}^{n+}}\to $ reactant ion concentration.

 $ \text{Nernst equation, for half cell}\text{.} \\ {{E}_{\left( \frac{Mn+}{M} \right)}}=E_{\left( \frac{{{M}^{n+}}}{M} \right)}^{{}^\circ }-\dfrac{2.303RT}{nF}\log \dfrac{\left[ M\left( s \right) \right]}{\left[ {{M}^{n+}}\left( aq \right) \right]} \\ \text{For full cell,} \\ {{E}_{cell}}=E_{cell}^{{}^\circ }-\dfrac{2.303RT}{nF}\log \dfrac{\left[ P \right]}{\left[ R \right]}\,\,\,\,\, \\ \left( Where,\left[ P \right]\to \text{Product conc}\text{. ions, }\left[ R \right]\to \text{Reactant conc}\text{. ions} \right) \\ \text{Given equation,} \\ \text{Anode:- }S{{n}^{2+}}\to S{{n}^{4+}}+2{{e}^{-}} \\ \text{Cathode:- }C{{u}^{2+}}+2{{e}^{-}}\to Cu \\ \text{Number of electron, }n=2 \\ {{E}_{cell}}=E_{cell}^{{}^\circ }-\dfrac{RT}{nF}\ln \dfrac{\left[ S{{n}^{4+}} \right]}{\left[ S{{n}^{2+}} \right]\left[ C{{u}^{2+}} \right]}\,\,\,\,\, \\ \left( \text{Because }Cu,\text{ activity found is one when it is solid state} \right)$

  In the reaction given below $ \\ Cu+2A{{g}^{+}}\to C{{u}^{2+}}+2Ag$ which of the statements is correct? $ \\ $ $(i)$ Oxidation occurs at cathode, $ \\ $ $(ii)$ Reduction occurs at cathode, $ \\ $ $(iii)$ Cathode is positive terminal of cell

  Since electrons flow from anode to cathode, the anode is the $-ve$ terminal and the cathode is the $+ve$ terminal of the cell.

 For the cell given, $ \\ Cu+2A{{g}^{+}}\to C{{u}^{2+}}+2Ag \\ $ Oxidation half reaction is: $ \\ $ Anode: $Cu\to C{{u}^{2+}}+2{{e}^{-}} \\ $ Reduction half reaction is: $ \\ $ Cathode: $2A{{g}^{+}}+2{{e}^{-}}\to 2Ag \\ $ Number of electron is $n=2$. $ \\ $ As the cathode of an electrochemical cell is identified by reduction and positive terminal of the cell. So, the statement $\left( \text{ii} \right)$ and $\left( \text{iii} \right)$ are correct. But oxidation occurs at anode, so the statement $\left( \text{i} \right)$ is incorrect.

 Calculate the standard Gibbs energy for the given reaction. $ \\ Zn\left| Z{{n}^{2+}}\left( aq \right) \right|\left| C{{u}^{2+}}\left( aq \right) \right|Cu\text{ for }E_{cell}^{o}=2.10Volt$

 Relation between Gibbs energy and electrode potential of a cell is ${{\Delta }_{r}}{{G}^{o}}=-nFE_{cell}^{o}$. Here, ${{\Delta }_{r}}{{G}^{o}}$ which stands for Gibbs energy of the cell reaction.

  $ Given, E_{cell}^{o}=2.10V \\ \text{Using,} \\ {{\Delta }_{r}}{{G}^{o}}=-nFE_{cell}^{o}\,\,\,\,\,...(i) \\ \text{In the reaction of cell given above,} \\ Zn\left( s \right)+C{{u}^{2+}}\left( aq \right)\to Cu\left( s \right)+Z{{n}^{2+}}\left( aq \right) \\ \text{Anode:- }Zn\to Z{{n}^{2+}}+2{{e}^{-}} \\ \text{Cathode:- }C{{u}^{2+}}+2{{e}^{-}}\to Cu \\ \text{No of electron is, }n=2 \\ \text{and }F=96500C\text{ (}\therefore \text{ standard value of F)} \\ \text{Substituting the values in equation }\left( \text{i} \right)\text{ we get,} \\ {{\Delta }_{r}}{{G}^{o}}=-2\times 96500\times 2.10 \\ \Rightarrow {{\Delta }_{r}}{{G}^{o}}=405300\text{Joule} \\ \text{Or }{{\Delta }_{r}}{{G}^{o}}=405.3KJ $

 In a Galvanic cell for $E_{cell}^{o}=0.18V$ at $298K$ and $n=2$ mole. What will be the value of equilibrium constant of the cell reaction.

 Using formula ${{\Delta }_{r}}{{G}^{o}}=-2.303RT\log {{K}_{C}};F=96500C;R=8.314\times {{10}^{-3}}$.

$ \text{Given, }E_{cell}^{o}=0.18V\text{ at }298K \\ \text{And, }n=2 \\ \text{Also, }F=96500C/mo{{l}^{-1}} \\ \text{Then, }{{\Delta }_{r}}{{G}^{o}}=-nFE_{cell}^{o} \\ \Rightarrow {{\Delta }_{r}}{{G}^{o}}=-2\times 96500\times 0.18 \\ \Rightarrow {{\Delta }_{r}}{{G}^{o}}=-34740J \\ \Rightarrow {{\Delta }_{r}}{{G}^{o}}=-34.74KJ \\ \text{Now,} \\ \text{Equilibrium constant }\left( {{K}_{C}} \right)\text{ we have,} \\ {{\Delta }_{r}}{{G}^{o}}=-2.303RT\log {{K}_{C}} \\ i.e.\text{ }log{{K}_{C}}=\dfrac{{{\Delta }_{r}}{{G}^{o}}}{-2.303\times R\times T} \\ \Rightarrow log{{K}_{C}}=\dfrac{-34.74}{-2.303\times 8.314\times {{10}^{-3}}\times 298} \\ \Rightarrow log{{K}_{C}}=\dfrac{-34.74}{-5.7}=6.09 \\ \Rightarrow {{K}_{C}}=Anti\log \left( 6.09 \right) $