For the following electrochemical cell at $289 K,$
$Pt (s) | H_{2} (g, 1 bar) | H^{+} (aq, 1 M) | | M^{4 +} (aq .) {, M}^{2+} (aq .) | Pt (s)$
$E_{c e l l} = 0.092 V$ when $\frac{[M^{2 +} (a q .)]}{[M^{4 +} (a q .)]} = 10^{x}$
Given: $E_{M^{4 +} / M^{2 +}}^{0} = 0.151 V; 2.303 \frac{R T}{F} = 0.059$
The value of $x$ is -

Question

For the following electrochemical cell at

289 K,

Pt (s) | H_{2} (g, 1 bar) | H^{+} (aq, 1 M) | | M^{4 +} (aq .) {, M}^{2+} (aq .) | Pt (s)

E_{c e l l} = 0.092 V

when

\frac{[M^{2 +} (a q .)]}{[M^{4 +} (a q .)]} = 10^{x}

Given:

E_{M^{4 +} / M^{2 +}}^{0} = 0.151 V; 2.303 \frac{R T}{F} = 0.059

The value of

x

is -

Aicharya · Accepted Answer

At anode : H 2 g ⇌ 2 H + a q + 2 e - At cathode : M 4 + a q + 2 e - ⇌ M 2 + a q Net cell reaction : H 2 g + M 4 + a q ⇌ 2 H + a q + M 2 + ( a q ) E c e l l = E c e l l 0 − 0.059 2 log [ M 2 + ] [ H + ] 2 [ M 4 + ] ( P H 2 ) Now, E c e l l = E M 4 + / M 2 + 0 - E H + / H 2 0 - 0.059 n . log ⁡ H + 2 M 2 + P H 2 . M 4 + 0.092 = 0.151 - 0 - 0.059 2 . log ⁡ 1 2 × M 2 + 1 × M 4 + ∴ M 2 + M 4 + = 10 2 ⇒ x = 2

For the following electrochemical cell at 289 K, Pt ( s ) | H 2 ( g , 1  bar )   | H + ( aq ,…

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For the following electrochemical cell at 289 K, Pt ( s ) | H 2 ( g , 1 bar ) | H + ( aq ,…