A cell in which the following reaction takes place- Mg s - 2Ag - 0-0001 M - Mg - 2 0-120M - 2Ag s if E- cell - 3-17 voltthen calculate Ecell

The correct option is B 2.96 volt E_cell= E^0_cell 0.05912 log Mg^2+(Ag^+)^2 E_cell = 3.17 0.0295 * log 0.120.0000 ...

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A cell in whi...

Question

A cell in which the following reaction takes place:-
$M g_{(s)} + 2 A g^{+} (0.0001 M) \to M g^{+ 2} (0.120 M) + 2 A g (s) if E_{c e l l}^{\circ} = 3.17 v o l t$
then calculate $E_{c e l l}$

2.0 volt

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2.96 volt

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3.50 volt

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3.17 volt

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Solution

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Similar questions

M g_{(} s) + 2 A g^{+} (0.0001 M) ⇌ M g^{2 +} (0.130 M) + 2 A g_{(s)}

E_{(c e l l)} E_{(c e l l)}^{0} = 3.17 v o l t

M g (s) + 2 A g^{+} (0.0001 M) \to M g^{2 +} (0.130 M) + 2 A g (s)

E_{(c e l l)} if E_{(c e l l)}^{o} = 3.17 V

M g + 2 {A g}^{+} (0.0001 M) \to {M g}^{2 +} (0.01 M) + 2 A g

E_{c e l l}^{o} = 3.177 V

E_{c e l l}

M g_{(s)} + 2 A g_{(a q)}^{+} \to M g_{(a q)}^{2 +} + 2 A g_{(s)}

E_{M g^{2 +} / M g}^{0} = - 2.37 V

E_{A g^{+} / A g}^{0} = 0.80 V

[M g^{2 +}] = 0.001 M; [A g^{+}] = 0.0001 M

Ni (s) + 2 A g^{+} (0.002 M) \to N i^{2 +} (0.160 M) + 2 Ag (s)

E_{(cell)}^{⊙} = 1.05 V

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