Question

Find the emf of the given cell.$Ni\left(s\right)|N{i}^{2+}(aq,1.0M)||A{u}^{3+}(aq,0.1M)|Au\left(s\right)$
$[E^0\text{for}N{i}^{2+}/Ni=-0.25V$
$E^0\text{for}A{u}^{3+}/Au=1.50V]$

• A. $1.25V$
• B. $-0.96V$
• C. $2.31V$
• D. $1.73V$

Answer 1

The correct option is D $1.73V$

For the given cell representation,
$Ni/N{i}^{2+}$ act as anode
$A{u}^{3+}/Au$ act as cathode

The half cell reactions are
$Ni\left(s\right)\to N{i}^{2+}\left(aq\right)+2e^{-}$
$A{u}^{3+}\left(aq\right)+3e^{-}\to Au\left(s\right)$

The cell reaction is
$3Ni\left(s\right)+2A{u}^{3+}\left(aq\right)\to 2Au\left(s\right)+3N{i}^{2+}\left(aq\right)$
Six electrons are involved in this reaction.
Thus, $n=6$

$E_{cell}^0=\text{SRP of substance reduced}-\text{SRP of substance oxidised}$
(SRP - Standard reduction potential)

$E_{cell}^0=E_{cathode\left(red\right)}^0-E_{anode\left(red\right)}^0$
$E_{cell}^0=1.50-(-0.25)$
$E_{cell}^0=1.75V$

By Nernst equation,
$E_{cell}=E^0-\frac{0.059}{n}\text{log}\left[\frac{(N{i}^{2+}{)}^3}{(A{u}^{3+}{)}^2}\right]$

$E_{\text{cell}}=1.75-\frac{0.059}{6}\text{log}\frac{1}{0.01}$

$E_{\text{cell}}=1.75-\frac{0.059}{6}\text{log}\left[100\right]$

$E_{\text{cell}}=1.75-0.02$
$E_{\text{cell}}=1.73V$