battery was made with a nickel metal electrode dipped into a 1.00 MNiC,HO)o f cl

Question

battery was made with a nickel metal electrode dipped into a 1.00 MNiC,HO)o f cl, and an gold metal electrode dipped in a 1.00 M AuCl, solution in the other 2H,02h sol ution in one Write the reduction half-reaction for this battery b) Write the oxidation half-reaction for this battery. c) Write the balanced redox equation for this battery d) What is the standard cell voltage (Ee) for this battery? e) If the concentration of Ni(C2H,02h remains at 1.00 M, what concentration of AuCls would result in a cell voltage of 1.00 volts?

Explanation / Answer

a)

reduction is gain in electron so

Au+3 + 3e- --> Au(s)

b)

oxidation i sloss of e- therefore

Ni2+ + 2e- --> Ni(s)

but msut be inverted sinc eit is oxidized

Ni(s) --> Ni2+ + 2e-

c)

balanced reaction (2e- anmd 3e- must be balanced)

2Au+3 + 6e- --> 2Au(s)

3Ni(s) --> 3Ni2+ + 6e-

add all

3Ni(s) + 2Au+3 + 6e- --> 2Au(s) + 3Ni2+ + 6e-

3Ni(s) + 2Au+3 --> 2Au(s) + 3Ni2+(aq)

d)

E°cell = Egold - Enickel = 1.52 --0.25 = 1.77 V

e)

Apply Nernst equation

E = E° - RT/(nF) * ln (Q)

Q = [Ni+2]^3 / [Au+3]^2

n = 2x3 = 6 e-,. F = 96500, R = 8.31 4J/molK, T = 298K

E = 1, E° = 1.77

substitute

E = E° - RT/(nF) * ln (Q)

1 = 1.77 - 8.314*298/(6*96500) * ln( [Ni+2]^3 / [Au+3]^2)

(1-1.77 )/(-8.314*298) * (6*96500) = ln( 1^3 / [Au+3]^2)

179.94 =  ln( 1^3 / [Au+3]^2)

[Au+3]^2 = 1 / exp(179.94 )

[Au+3] = (7.129*10^-79)^0.5 = 8.44*10^-40

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