Value of equilibrium constant for the reaction, Cu2+(aq) +Sn2+(aq) → Sn4+(aq) +Cu(s) is (Given E°Cu2+/Cu = 0.34V, E°Sn4+/Sn2+ =0.15V)

Question

Value of equilibrium constant for the reaction, Cu2+(aq) +Sn2+(aq) → Sn4+(aq) +Cu(s) is (Given E°Cu2+/Cu = 0.34V, E°Sn4+/Sn2+ =0.15V) (A) 1.64 × 103 (B) 3.82 × 1016 (C) 2.69 × 106 (D) 7.23 × 1012

Tardigrade Admin · Accepted Answer

E° textcell =E° textcathode -E° textanode = 0.34 -0.15 =0.19 V Applying, E° textcell = (0.0591/n) log K log Kc = (0.19 ×2/0.0591) =6.43 Taking antilog, Kc = 2.69 ×106

Q. Value of equilibrium constant for the reaction,
$C u_{(a q)}^{2 +} + S n_{(a q)}^{2 +} \to S n_{(a q)}^{4 +} + C u_{(s)}$ is (Given $E_{C u^{2 +} / C u}^{\circ} = 0.34 V, E_{S n^{4 +} / S n^{2 +}}^{\circ} = 0.15 V$ )

$1.64 \times 1 0^{3}$

$3.82 \times 1 0^{16}$

$2.69 \times 1 0^{6}$

$7.23 \times 1 0^{12}$

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.34 - 0.15 = 0.19 V$
Applying, $E_{cell}^{\circ} = \frac{0.0591}{n}$ log $K$
log $K_{c} = \frac{0.19 \times 2}{0.0591} = 6.43$
Taking antilog, $K_{c} = 2.69 \times 1 0^{6}$

Q. Value of equilibrium constant for the reaction, Cu(aq)2+​+Sn(aq)2+​→Sn(aq)4+​+Cu(s)​ is (Given ECu2+/Cu∘​=0.34V,ESn4+/Sn2+∘​=0.15V)

1.64×103

3.82×1016

2.69×106

7.23×1012