Given : $E^{0} (C u^{2 +} / C u)$ =0.337 V and $E^{0} (S n^{2 +} / S n)$ =-0.136 V. Which of the following statements is correct?

C u^{2 +}

ions can be reduced by

H_{2}

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Cu can be oxidised by

H_{2}

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S n^{2 +}

ions can be reduced by

H_{2}

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Cu can reduce

S n^{2 +}

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Solution

The correct option is A $C u^{2 +}$ ions can be reduced by $H_{2}$
Compare the reduction potentials.
$E_{(\frac{C u^{2 +}}{C u})}^{\circ} > E_{(\frac{S n^{2 +}}{S n})}^{\circ}$
$∴ C u$ electode act as cathode
$s n$ electrode act as anode.
Therefore $C u^{2 +}$ ion can be reduced by $H_{2}$ .

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Similar questions

E_{{C u}^{2 +} | C u}^{\circ} = 0.34 V

E_{S n^{2 +} | S n}^{\circ} = - 0.136, E_{H^{+} | H_{2}}^{\circ} = - 0.0 V

$E_{A g^{+} | A g}^{\circ} = 0.8$ volt and $E_{Z n^{2 +} | Z n}^{\circ} = - 0.76$ volt

Which of the following is correct?

CuO + H_{2} \to Cu + H_{2} O

For the reaction $C u O + H_{2} \to C u + H_{2} O$
Which of the following statement is correct?

E_{C u^{2 +} | C u}^{\circ} = 0.34 V

E_{S n^{2 +} | S n}^{\circ} = - 0.136 V

E_{H^{+} | H_{2}}^{\circ} = - 0.0 V

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