Question

To find the standard potential of $M{3}^{+}/M$ electrode, the following cell is constituted : $Pt/M/M^{3+}\left(0.001mol{L}^{-1}\right)/A{g}^{+}\left(0.01mol{L}^{-1}\right)/Ag$
The emf of the cell is found to be 0.421 volt at 298 K The standard potential of half reaction $M^3+3e^{-}\to M$ at 298 K will be :
(Given $E_{A{g}^{+}/Ag}^e$ at 298 K =0.80 Volt)

• A. 0.32 Volt
• B. 0.66 Volt
• C. 1.28 Volt
• D. 0.38 Volt

Answer 1

The correct option is A 0.32 Volt

$E_{cell}=E_{cell}^0-\frac{0.0592}{n}log\frac{\left[M^{3+}\right]}{[A{g}^{+}{]}^3}$
$0.421=E_{cell}^0-\frac{0.0592}{3}log\frac{0.001}{(0.01{)}^3}$
$E_{cell}^0=0.48V$
$E_{cell}^0=E_{Ag}^0-E_M^0$
$0.48=0.80-E_{Ag}^0$
$E_{Ag}^0=0.32$ V
The standard potential of half reaction $M^3+3e^{-}\to M$ at 298 K at 298 K will be 0.32 V.