For the following electrochemical cell at 298 K- Pt s - H - 2 g - 1 - bar - H - aq - 1 M M -4- aq - M -2- aq - Pt s 1E cell -0-092 - V - when -frac-M -2-a q-M -4-a q-10- X lGiven - E - 0- M -4- - M -2-0-151 - V - 2-303 frac RT F -0-059 - V The value of x is-A- -2B- -1C- 1D- 2

The correct option is D 2Anode : H_2 nbsp; ⇌ 2H^+ + 2e Cathode : M^4+ + 2e ⇌ M^2+ Net cell reaction: H_2 + M^4+⇌ 2H^+ + M^2+ E_cell = E^0_cell 0.059/2 lo ...

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Byju's Answer

Standard XII

Chemistry

Gibb's Energy and Nernst Equation

For the follo...

Question

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

-2

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Solution

The correct option is D 2
$A n o d e : H_{2} ⇌ 2 H^{+} + 2 e C a t h o d e : M^{4 +} + 2 e ⇌ M^{2 +}$
Net cell reaction: $H_{2} + M^{4 +} ⇌ 2 H^{+} + M^{2 +} E_{c e l l} = E_{c e l l}^{0} - \frac{0.059}{2} l o g \frac{[H^{+}] [M^{2 +}]}{[M^{4 +}] \times P_{H_{2}}} 0.092 = 0.151 - \frac{0.059}{2} l o g 10^{x} 0.092 = 0.151 - \frac{0.059}{2} x \frac{0.059 x}{2} = 0.151 - 0.092 0.059 x = 0.059 \times 2 x = 2$

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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 +}}^{\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:

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. If

E_{C u | C u^{+}}^{0} = - 0.53 V

and

E_{C u^{+} | C u^{2 +}}^{0} = - 0.15 V

For given cell,

P t (s) | H_{2} (g, 101325 p a s c a l) | H_{2} S O_{4} (a q .) | | C u^{2 +} (a q ., 1 M) | C u (s)

E_{c e l l} = 0.34 V a t 298 K

Then, the molarity of

H_{2} S O_{4} (a q .)

solution is equal to:

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

The correct option is D 2Anode:H2⇌2H++2eCathode:M4++2e⇌M2+ Net cell reaction: H2+M4+⇌2H++M2+Ecell=E0cell−0.0592log[H+][M2+][M4+]×PH20.092=0.151−0.0592log 10x0.092=0.151−0.0592x0.059 x2=0.151−0.0920.059 x=0.059×2x=2