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 E cell -0-092 V when - M 2- aq - M 4- aq -10 xGiven - E M 4- - M 2-0-151 V - 2-303 RT - F -0-059 VThe value of x isA- -2B- 2C- 1D- -1

The correct option is B 2Cell reaction M^4+ + H_2 → M^2+ + 2H^⊕ E_cell = E^∘_cell 0.06/2 log [M^2+]/[M^4+] 0.092 = 0.151 0.03 log M^2+/M^4+ = 0.06 = ...

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Byju's Answer

Standard VII

Chemistry

Nersnt Equation

For the follo...

Question

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

-2

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-1

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Solution

The correct option is D 2
Cell reaction
$M^{4 +} + H_{2} \to M^{2 +} + 2 H^{\oplus} E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.06}{2} l o g \frac{[M^{2 +}]}{[M^{4 +}]} 0.092 = 0.151 - 0.03 l o g \frac{M^{2 +}}{M^{4 +}} = - 0.06 = - 0.03 l o g_{10} \frac{[M^{+ 2}]}{[M^{+ 4}]} \frac{[M^{2 +}]}{[M^{4 +}]} = 10^{2} [G i v e n \frac{M^{2 +} (a q)}{[M^{4 +} (a q)]}] 10^{x} = 10^{2} = x = 2$

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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 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 +}}^{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 +}}^{0} = 0.151 V; 2.303 \frac{R T}{F} = 0.059 V

The value of x is:

Q. For the given electrochemical cells :

Cell 1:

E_{F e^{2 +} (a q) / F e (s)}^{\circ} = - 0.44 V

E_{H^{+} (a q) / H_{2} (g)}^{\circ} = 0 V

Cell 2:

E_{M g^{2 +} (a q) / M g (s)}^{\circ} = - 2.37 V

E_{H^{+} (a q) / H_{2} (g)}^{\circ} = 0 V

Choose the correct option:

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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=0.092V when[M2++(aq)][M4+(aq)]=10x Given:E∘M4+/M2+=0.151V;2.303RTF=0.059V The value of x is

The correct option is D 2Cell reaction M4++H2→M2++2H⊕Ecell=E∘cell−0.062log[M2+][M4+]0.092=0.151−0.03logM2+M4+=−0.06=−0.03log10[M+2][M+4][M2+][M4+]=102 [GivenM2+(aq)[M4+(aq)]]10x=102=x=2