23 Given the following two half reactions: Cd^2+ (aq) + 2e^- rightarrow Cd(s) E

Question

23

Given the following two half reactions: Cd^2+ (aq) + 2e^- rightarrow Cd(s) E degree = -0.40 V Z^4+ + 4e^rightarrow Zr(s) E degree = -1.53 V determine E degree and the spontaneity of the following reaction: 2Cd^2+ (aq) + Zr(s) rightarrow 2Cd(s) + Zr^4-(aq) +1.13 V and not spontaneous +1.13 V and spontaneous -1.13 V and not spontaneous -1.13 V and spontaneous not enough information given Calculate Delta G degree at 25 degree C for a cell reaction where three moles of electrons are transferred and E degree -3618 kJ -201 kJ -145 kJ -23 kJ

Explanation / Answer

AS Zr has the lowest potential, this means that this is the reducting agent, and the Cd is the oxidizing agente, therefore Zr is being oxidated and Cd is reducted. The half reactions are:

(Cd2+ + 2e- = Cd) *2
Zr = Zr4+ + 4e-

2Cd2+ + 4e- = 2Cd
Zr = Zr4+ + 4e-

2Cd2+ + Zr = 2Cd + Zr4+
Eº = Eºred - Eºoxi
Eº = -0.40 + 1.53 = +1.13 V This means that the reaction is spontaneous.

Hope this helps

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