An electrochemical cell is constructed with an open switch as shown below : When the switch is closed, mass of tin-electrode increase. If Eº (Sn2+ / Sn) = – 0.14 V and for Eº (Xn+ / X) = –0.78 V and initial emf of the cell is 0.65 V, determine n and indicate the direction of electron flow in the external circuit.

Engineering

Chemistry

Faraday Law

Question

An electrochemical cell is constructed with an open switch as shown below :

When the switch is closed, mass of tin-electrode increase. If Eº (Sn²⁺ / Sn) = – 0.14 V and for Eº (Xⁿ⁺ / X) = –0.78 V and initial emf of the cell is 0.65 V, determine n and indicate the direction of electron flow in the external circuit.

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Since mass of Sn increasing, Sn - electrode is working as cathode and X - metal electrode anode and electrons are flowing from X-electrode to Sn-electrode in the external circuit.

$0.65 = E_{oxid} + E_{red} = {0.78 - \frac{0.0591}{n} \log (0.1)} + {0 - 0.14 - \frac{0.0591}{2} \log \frac{1}{0.5}}$

$0.01 = - \frac{0.0591}{n} \times (- 1) - \frac{0.0591}{2} \times 0.301 = 0.0591 (\frac{1}{n} - \frac{0.301}{2}) n = 3$

Electrons flow from X electrode to Zn electrode.

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