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:

1 M

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0.5 M

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2 M

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0.25 M

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Solution

The correct option is B $0.5 M$
From the given data,
$E_{C u^{2 +} | C u}^{0} = 0.34 V$

Nernst equation:
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{2} l o g [H^{+}]^{2}$

$0.34 = 0.34 - \frac{0.0591}{2} l o g [H^{+}]^{2}$

$[H^{+}] = 1$ moles per litre

So, molarity of $H_{2} S O_{4} = 0.5 M$

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Similar questions

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:

In the electrochemical cell $P t (s) | H_{2} (g), 1 a t m | H^{+} (1 M) | | C u^{2 +} (1 M) | C u (s)$ , which one of the following statements is true?

Q. Calculate the equilibrium constant of the reaction

F e + C u S O_{4} ⇌ F e S O_{4} + C u

25^{o} C

Given

E^{0} [F e \to F e^{2 +}) = 0.44 V

and

E^{0} (C u \to C u^{2 +}) = - 0.337 V

Q. Calculate the equilibrium constant for the reaction

F e + C u S O_{4} ⇌ F e S O_{4} + C u

25^{0}

C. Given

E^{0} (F e / F e^{2 +}) = 0.44 V, E^{0} (C u / C u^{2 +}) = - 0.337 V

Q. Calculate the equilibrium constant for the reaction:

F e + C u S O_{4} ⟶ F e S O_{4} + C u

25^{o}

[Given :

E_{(F e / F e^{2 +})}^{0} = 0.44 V

;

E_{(C u / C u^{2 +})}^{0} = - 0.337 V

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If E0Cu|Cu+=−0.53 V and E0Cu+|Cu2+=−0.15 V For given cell, Pt(s)|H2(g,101325 pascal)|H2SO4(aq.)||Cu2+(aq., 1M)|Cu(s) Ecell=0.34V at 298K Then, the molarity of H2SO4(aq.) solution is equal to:

The correct option is B 0.5 MFrom the given data, E0Cu2+|Cu=0.34 V Nernst equation: Ecell=E0cell−0.05912log[H+]2 0.34=0.34−0.05912log[H+]2 [H+] = 1 moles per litre So, molarity of H2SO4=0.5 M

The correct option is B $0.5 M$
From the given data,
$E_{C u^{2 +} | C u}^{0} = 0.34 V$

Nernst equation:
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{2} l o g [H^{+}]^{2}$

$0.34 = 0.34 - \frac{0.0591}{2} l o g [H^{+}]^{2}$

$[H^{+}] = 1$ moles per litre

So, molarity of $H_{2} S O_{4} = 0.5 M$