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Enthalpy

The standard ...

Question

The standard free energy change for the following reactions is $- 210$ kJ. What is the standard cell potential?
$2 H_{2} O_{2} (a q) \to 2 H_{2} O (l) + O_{2} (g)$

+ 0.752

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+ 1.09

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+ 0.420

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+ 0.640

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Solution

The correct option is C $+ 1.09$ V
${2 H}_{2} O_{2} (a q) ⟶ 2 H_{2} O (l) + O_{2} (g)$
According to the formula,
${- Δ G}^{0} = n F E_{c e l l}^{0}$
$E_{c e l l}^{0} = ?$
$n = 2$ , $F = 96500$ $C$
$E_{c e l l}^{0} = \frac{210 \times 10^{3}}{2 \times 96500}$
$E_{c e l l}^{0} = 1.09$ V

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Similar questions

2 H_{2} O_{2} (a q) \to 2 H_{2} O (l) + O_{2} (g)

{Z n}^{2 +} | Z n

- 0.76 V

{C u}^{2 +} | C u

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(V)

(k J {m o l}^{- 1})

3 F e^{2 +} (a q) + 2 C r (s) = 2 C r^{3 +} (a q) + 3 F e (s)

E_{F e^{2 +} / F e}^{o} = - 0.44

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(F = 96500 C)

The standard reaction gibbs energy for the electrochemical reaction.

$C r_{2} O_{7}^{2 -} + 2 F e + 14 H^{+} \to 2 C r^{3 +} + 2 F e^{3 +} + 2 H_{2} O$ is $- 793 k J m o l^{- 1}$ . The standard cell potential is:

C r_{2} O_{7}^{2 -} + 6 F e^{2 +} + 14 H^{+} \to 2 C r^{3 +} + 6 F e^{3 +} + 7 H_{2} O i s - 793 k J m o l^{- 1}

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