Question

Calculate the valency of metal from the following cell at $298K$, if the emf of the cell is $0.074V$
$M|M^{n+}(aq,0.0083M)||M^{n+}(aq,0.148M)|M$

Take:
$log\left(0.056\right)=-1.25$

• A. 1
• B. 2
• C. 4
• D. 3

Answer 1

The correct option is A 1

$M|M^{n+}(aq,0.0083M)||M^{n+}(aq,0.148M)|M$

Life side electrode will act as anode and right side electrode will act as cathode.

For the given concentration cell,

$E=E^0-\frac{0.0591}{n}log\frac{\left[Anode\right]}{\left[Cathode\right]}$

For concentration cell,
$E^0=0$
$\therefore$
$E=-\frac{0.0591}{n}log\frac{\left[Anode\right]}{\left[Cathode\right]}$

$0.074=-\frac{0.0591}{n}log\frac{\left[0.0083\right]}{\left[0.148\right]}$

$0.074=-\frac{0.0591}{n}log0.056$

$0.074=\frac{0.0591}{n}\times 1.25$

$n=\frac{0.0591}{0.074}\times 1.25$

$n=0.998$

$n\approx 1$