For the following electrochemical cell at 298 K- Pts -H2 g-1 bar-H- aq-1M- - M4- aq-M2- aq-PtsEcell-0-092 V when-M2-aq-M4-aq-10xGiven-EM4-M2-0-0-151 V- 2-303RTF-0-059 VThen value of x is -

At Anode : H_2 (s)→2H^+ +2e^– At cathode :M^4++2e^ → M^2+ Adding both the reactions M^4++H_2→ M^2++2H^+ According to Nernst equation, E=E^0 0.059nlog_10([M^2+][ ...

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Byju's Answer

Standard VII

Chemistry

Nersnt Equation

For the follo...

Question

For the following electrochemical cell at $298 K$ ,
$P t (s) | H_{2} (g, 1 b a r) | | (H^{+} (a q .1 M) | | M^{4 +} (a q), M^{2 +} (a q)) | P t (s)$
$E_{c e l l} = 0.092 V when \frac{[M^{2 +} (a q)]}{[M^{4 +} (a q)]} = 10^{x}$
Given: $E_{M^{4} / M^{2 +}}^{0} = 0.151 V; 2.303 \frac{R T}{F} = 0.059 V$
Then value of $x$ is :

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Solution

The correct option is A $2$
$At Anode : H_{2} (s) \to 2 H^{+} + 2 e^{-}$
$At cathodeode : M^{4 +} + 2 e^{-} \to M n^{2 +} Adding both the reactions M^{4 +} + H_{2} \to M n^{2 +} + 2 H^{+}$
According to Nernst equation,
$E = E^{0} - \frac{0.059}{n} {log}_{10} (\frac{[M n^{2 +}] [H^{+}]^{2}}{[M n^{4 +}] P_{H_{2}}})$
$\Rightarrow 0.092 = 0.151 - \frac{0.059}{2} {log}_{10} (10^{x})$
$\Rightarrow 0.092 = 0.151 - \frac{0.059}{2} x$
$\Rightarrow x = 2$

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Similar questions

Q. For the following electrochemical cell at 298 K,

P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q, 1 M) | | M^{4 +} (a q), M^{2 +} (a q) | P t (s) E_{c e l l} = 0.092 V w h e n \frac{[M^{2 +} (a q)]}{[M^{4 +} (a q)]} = 10^{x} G i v e n : E_{M^{4 +} / M^{2 +}}^{0} = 0.151 V; 2.303 \frac{R T}{F} = 0.059 V

The value of x is:

Q. For the following electrochemical cell at 298 K,

P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q, 1 M) | | M^{4 +} (a q), M^{2 +} (a q) | P t (s)

E_{c e l l} = 0.092 V w h e n \frac{[M^{2 +} + (a q)]}{[M^{4 +} (a q)]} = 10^{x}

G i v e n : E_{M^{4 +} / M^{2 +}}^{\circ} = 0.151 V; 2.303 \frac{R T}{F} = 0.059 V

The value of x is

Q. For the following electrochemical cell at 298K,

P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q, 1 M) | | M^{4 +} (a q), M^{2 +} (a q) | P t (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 V

The value of x is :

Q. For the following electrochemical cell at 298 K,

P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q, 1 M) | | M^{4 +} (a q), M^{2 +} (a q) | P t (s)

E_{c e l l} = 0.092 V w h e n \frac{[M^{2 +} + (a q)]}{[M^{4 +} (a q)]} = 10^{x}

G i v e n : E_{M^{4 +} / M^{2 +}}^{\circ} = 0.151 V; 2.303 \frac{R T}{F} = 0.059 V

The value of x is

Q. For the following electrochemical cell at 298 K,

P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q, 1 M) | | M^{4 +} (a q), M^{2 +} (a q) | P t (s) E_{c e l l} = 0.092 V w h e n \frac{[M^{2 +} (a q)]}{[M^{4 +} (a q)]} = 10^{x} G i v e n : E_{M^{4 +} / M^{2 +}}^{0} = 0.151 V; 2.303 \frac{R T}{F} = 0.059 V

The value of x is:

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For the following electrochemical cell at 298 K, Pt(s)|H2(g,1bar)||(H+(aq.1M)||M4+(aq),M2+(aq))|Pt(s) Ecell=0.092 V when[M2+(aq)][M4+(aq)]=10x Given:E0M4/M2+=0.151 V; 2.303RTF=0.059 V Then value of x is :

The correct option is A 2At Anode:H2(s)→2H++2e– At cathodeode:M4++2e−→Mn2+Adding both the reactionsM4++H2→Mn2++2H+ According to Nernst equation, E=E0−0.059nlog10([Mn2+][H+]2[Mn4+]PH2) ⇒0.092=0.151−0.0592log10(10x) ⇒0.092=0.151−0.0592x ⇒x=2