Calculate the standard cell potential (in V) of the cell in which the following reaction takes place:
${Fe}^{2 +}\left(aq\right) + A{g}^{+}\left(aq\right) \to F{e}^{3 +}\left(aq\right) + Ag ( s )$
Given that
${E}_{{Ag}^{+} / Ag}^{o} = xV$
${E}_{{Fe}^{2 +} / Fe}^{o} = yV$
${E}_{{Fe}^{3 +} / Fe}^{o} = zV$

Question

Calculate the standard cell potential (in V) of the cell in which the following reaction takes place:

${Fe}^{2 +}\left(aq\right) + A{g}^{+}\left(aq\right) \to F{e}^{3 +}\left(aq\right) + Ag ( s )$

Given that
${E}_{{Ag}^{+} / Ag}^{o} = xV$
${E}_{{Fe}^{2 +} / Fe}^{o} = yV$
${E}_{{Fe}^{3 +} / Fe}^{o} = zV$

Quantrex Academy · Accepted Answer

Correct answer: A. x + 2y - 3z

A F e ^ + 2 + C A g ^ + → F e ^ 3 + + Ag E _ cell ^ o = SRP of cathode - SRP of anode = E _ A g ^ + | Ag ^ o - E _ F e ^ + 3 | F e ^ + 2 ^ o = x - E _ F e ^ 3 + | F e ^ + 2 ^ o Since E ° cell is an intensive property, the E ° cell values are not additive by nature. But the standard Gibbs free energy change is an extensive property and hence can be added. (i) F e ^ + 3 → Fe + 3 e ^ - → Z = E _ 1 ^ o ; Δ G _ 1 = - 3 × F × z (ii) F e ^ + 2 → Fe + 2 e ^ - Y = E _ 2 ^ o ; Δ G _ 2 = - 2 × F × y Subtracting (ii) from (i) F e ^ + 3 → F e ^ + 2 Δ G _ 3 = Δ G _ 1 - Δ G _ 2 ( E ^ o = - 2y + 3z) E _ cell ^ o = x - (- 2y + 3z) = x + 2y - 3z

Calculate the standard cell potential (in V) of the cell in which...

Options

Correct answer

Step-by-step solution

More from Electrochemistry