The emf of the cell$\ {Mg|\ Mg}^{2 +}(0.01\ M)||{\ Sn}^{2 +}(0.1\ M)|Sn$ at 298 K is (Given, $E_{{Mg}^{2 +},\ Mg}^{{^\circ}} = \ - 2.34\ V, - 2.34\ V,E_{{Sn}^{2 +},\ \ Sn}^{{^\circ}} = \ - 0.14\ V$)

Question

Dr. Abubacker · Accepted Answer

(a)
Cell reaction is $Mg + {Sn}^{2 +} \rightarrow {Mg}^{2 +} + Sn$
$E_{cell} = E_{cell}^{{^\circ}} - \frac{0.0591}{2}\log\frac{\left\lbrack {Mg}^{2 +} \right\rbrack}{\left\lbrack Sn^{2 +} \right\rbrack}$
$= (2.34 - 0.14) - \frac{0.0591}{2}\log\frac{10^{- 2}}{10^{- 1}} = 2.23\ V$

Q. The emf of the cell$\ {Mg|\ Mg}^{2 +}(0.01\ M)||{\ Sn}^{2 +}(0.1\ M)|Sn$ at 298 K is (Given, $E_{{Mg}^{2 +},\ Mg}^{{^\circ}} = \ - 2.34\ V, - 2.34\ V,E_{{Sn}^{2 +},\ \ Sn}^{{^\circ}} = \ - 0.14\ V$)

Explanation

Reviewed by Dr. Abubacker

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