Question

$Pt\left(s\right)\mid M\left(s\right)\mid M^{3+}(aq,0.002mol{L}^{-1})\parallel A{g}^{+}(aq,0.01mol{L}^{-1})\mid Ag\left(s\right)$

The emf of the given cell is found to be $0.42V$ at $25{}^{\circ }C$. Calculate the standard reduction potential of the half reaction.
$M^{3+}\left(aq\right)+3e^{-}\to M\left(s\right)$

Given:${E^0}_{A{g}^{+}/Ag}=0.80V,\log 2=0.3$

Answer 1

From the cell representation,
silver electrode will act as cathode and the $M^{3+}\left(aq\right)/M\left(s\right)$ couple will act as anode.
$\therefore$
The cell reaction will be,
$M\left(s\right)+3A{g}^{+}\left(aq\right)\to 3Ag\left(s\right)+M^{3+}\left(aq\right)$

According to Nernst equation,
$E_{cell}=E_{cell}^0-\frac{0.0591}{3}log\frac{\left[M^{3+}\right]}{{\left[A{g}^{+}\right]}^3}$

$\Rightarrow 0.42=E_{cell}^0-\frac{0.0591}{3}{\log }_{10}\frac{\left(0.002\right)}{{\left(0.01\right)}^3}$

$\Rightarrow 0.42=E_{cell}^0-\frac{0.0591}{3}\log (2\times {10}^3)$

$\Rightarrow 0.42=E_{cell}^0-\frac{0.0591}{3}\times 0.3-0.0591$

$\Rightarrow 0.42=E_{cell}^0-0.065$

$\Rightarrow E_{cell}^0=0.42+0.065$

$\Rightarrow E_{cell}^0=0.485V$

$E_{cell}^0=E_{cathode}^o-E_{anode}^o$
$\Rightarrow E_{anode}^0=0.80-0.485$
$\Rightarrow E_{anode}^0=0.315V$

Thus, the standard reduction potential of the half cell reaction is $0.315V$.