Question

The emf of the cell $Ni$/$N{i^{ + 2}}$$(1.0M)$$\parallel A{u^{ + 3}}(0.1M)$/$Au[{E^o}$ for $N{i^{ + 2}}$/$Ni$$ = - 0.25V,{E^o}$ for $Au = 1.50V$$]$ is given as:
(A) $1.25V$
(B) $ - 1.75V$
(C) $1.75V$
(D) $1.73V$

Answer 1

Hint: Electromotive force or emf is basically the difference between the electrode potentials of the two electrodes constituting an electrochemical cell. It is determined with the help of Nernst equation.

Complete step by step solution:
As we know that Electromotive force or emf is basically the difference between the electrode potentials of the two electrodes constituting an electrochemical cell. It is also called cell potential of a cell when no current flows through it. It can be expressed as:
$E_{cell}^o = [standard\;reduction\;potential\;of\;cathode] - [standard\;reduction\;potential\;of\;anode]$
Or $E_{cell}^o = {E^o}(cathode) - {E^o}(anode)$
Let us first write the chemical equation to find out the emf,
$3Ni + 2A{u^{3 + }} \to 3N{i^{2 + }} + 2Au$
Now we know that emf as well as electrode potential can be determined by Nernst equation therefore using Nernst equation and putting all the given values we can calculate the emf of the cell:
$
  {E_{cell}} = {E^o} - \dfrac{{0.059}}{n}\log \dfrac{{{{(N{i^{2 + }})}^3}}}{{{{(A{u^{3 + }})}^2}}} \\
  {E_{cell}} = (1.50 - ( - 0.25)) - \dfrac{{0.059}}{6}\log [100] \\
  \Rightarrow {E_{cell}} = 1.75 - 0.02 \\
  \Rightarrow {E_{cell}} = 1.73V \\
 $
Therefore the correct answer is (D).

Note: We can also calculate the electromotive force or emf of the cell by knowing the standard electrode potential from the series as we know that $E_{cell}^o = {E^o}(right) - {E^o}(left)$. The reduction potential of various electrodes have been arranged in increasing order and oxidation is in decreasing order; that series is called electrochemical series or electromotive series.