Construct a cel from Ni(s), Ni^2+(l), Cd(s) and Cd^2+(l) a. What would be the an

Question

Construct a cel from Ni(s), Ni^2+(l), Cd(s) and Cd^2+(l) a. What would be the anode and what would be the cathode? b. indicate the half reactions happening at each the anode and the cathode. c. indicate which direction the electrons will flow. d. a battery is place between the anode and cathode to drive the reaction, what is the minimum voltage to drive the reaction? Given: Ni^2+(aq) + 2e- ---> Ni (s) = -0.23v Cd^2+ (aq) + 2e- -----> Cd(s) = -0.40v

Explanation / Answer

a) cd will be anode and Ni will be cathode b) Cd(s) ---->Cd^2+ (aq) + 2e- (at anode) Ni^2+(aq) + 2e- ---> Ni (s) ( at cathode) electron will flow from Cd to Ni minimum voltage to Drive reaction is 0.17V

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