If E° (cell) = 3.17 V for the cell in which the following reaction takes place: Mg(s) + 2Ag+(0.0001M) → Mg2+(0.130M) + 2Ag(s) then its E(cell) will be

Question

If E° (cell) = 3.17 V for the cell in which the following reaction takes place: Mg(s) + 2Ag+(0.0001M) → Mg2+(0.130M) + 2Ag(s) then its E(cell) will be (A) 3.56 V (B) 2.96 V (C) 5.35 V (D) 7.25 V

Tardigrade Admin · Accepted Answer

The cell can be written as Mg |Mg+ (0.130 M)||Ag+(0.0001 M) |Ag E(cell) = E°(cell) - (RT/2F) ln([Mg2+]/[Ag+]2) = 3.17 V - (0.059 V/2) log (0.130/(0.0001)2) = 3.17 V - 0.21 V = 2.96 V

Q. If $E_{(ce ll)}^{\circ} = 3.17 V$ for the cell in which the following reaction takes place: $M g_{(s)} + 2 A g^{+} (0.0001 M) \to M g^{2 +} (0.130 M) + 2 A g_{(s)}$
then its $E_{(ce ll)}$ will be

3.56 V

2.96 V

5.35 V

7.25 V

The cell can be written as
$M g ∣ M g^{+} (0.130 M) ∣∣ A g^{+} (0.0001 M) ∣ A g$
$E_{(ce ll)} = E_{(ce ll)}^{\circ} - \frac{RT}{2 F} l n \frac{[ M g ^{2 +} ]}{[ A g ^{+} ] ^{2}}$
$= 3.17 V - \frac{0.059 V}{2} l o g \frac{0.130}{( 0.0001 ) ^{2}}$
$= 3.17 V - 0.21 V = 2.96 V$

Q. If E(cell)∘​=3.17V for the cell in which the following reaction takes place: Mg(s)​+2Ag+(0.0001M)→Mg2+(0.130M)+2Ag(s)​ then its E(cell)​ will be