Assume that you build an electrochemical cell based on the following two half re

Question

Assume that you build an electrochemical cell based on the following two half reaction: Zn^2+ (aq)/Zn(s) and Ag^+/Ag(s). The standard reduction potential for Zn^2+ is -0.763 V and the standard reduction potential Ag^+ is 0.80 V. (a) Write a balanced equation for the product-favored reaction that occurs in the cell and calculate E- (b) Which electrode is the anode and which is the cathode? (c) Do electrons flow from the Zn electrode to the Ag electrode or vice versa? (d) If a salt bridge containing NaNO_3 connects the two half-cells, in which direction do the nitrate ions move, from the zinc to the silver compartment or vice versa? (c) Using the Nernst equation calculate the net cell potential if the following concentration of metals are present: [Zn^2+] = .35 M and [Ag^+] = .09 M

Explanation / Answer

Q1.

a)

The reaction from above shows oxidation/reduction so

Ag+(aq) + Zn(s) --> Ag(s) + Zn+2(aq)

balance

2Ag+(aq) + Zn(s) --> 2Ag(s) + Zn+2(aq)

E° = Ered - Eox = 0.80 --0.763 = 1.563

b)

electrodes:

anode -- >oxidation occurs hee... so this must be Zinc

cathode --> reductin occurs here, so this is silver

c)

electrons flow form SOLID zinc electrode to the auqeous solution in silver

d)

The NO3- ions must counterbaance the loss of + ions, which is Zn+2 + 2e- -> Zn(s)

so it goes to the zinc solution

e)

For E = E° -0.0592/n*log(Q)

Q = [Zn+2]/[Ag+]^2

n = 2electron

E° = 1.563

For E = E° -0.0592/n*log(Q)

For E = 1.563 -0.0592/2*log((0.35)/(0.09^2))

E = 1.514586 V

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