In a galvanic cell, one half-cell consists of a cobalt strip dipped into a 1.00

ID: 1059995 • Letter: I

Question

In a galvanic cell, one half-cell consists of a cobalt strip dipped into a 1.00 M solution of Co(NO3)2. In the second half-cell, solid tellurium is in contact with a 1.00 M solution of Te(NO3)4. Te is observed to plate out as the galvanic cell operates, and the initial cell voltage is measured to be 0.848 V at 25°C.

(a) Write balanced equations for the half-reactions at the anode and the cathode. Show electrons as e-. Use the smallest integer coefficients possible and the pull-down boxes to indicate states. If a box is not needed, leave it blank.

Half-reaction at anode (do not multiply by factor):

Half-reaction at cathode (do not multiply by factor):

(b) Calculate the standard reduction potential of a Te4+|Te half-cell. The standard reduction potential of the Co2+|Co electrode is -0.280 V.

V

Please write clear/concise step-by-step reasoning + answer,

many thanks, Max.

(aq)(s)(l)(g) + (aq)(s)(l)(g) (aq)(s)(l)(g) + (aq)(s)(l)(g)

Explanation / Answer

cathode --> Te+4(aq)+ 4e- --> Te(s) E = 0.568

anode --> Co2+(aq) + 2e– Co(s) –0.28

Note that

E°cell = Ered - Eox = 0.568 - –0.28 = 0.848V

which is what it marked

Find Potential for...

cathode --> Te+4(aq)+ 4e- --> Te(s) E = x

anode --> Co2+(aq) + 2e– Co(s) –0.28

E°cell = Ecathode - Eanode

if cell marked

0.848 then

0.848 = Ecathode - -0.28 V

Ecathode = 0.848 -0.28 = 0.568 V

So for Te+4 --> Te(s) we must have a reduction potential of 0.568 V

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In a galvanic cell, one half-cell consists of a cobalt strip dipped into a 1.00

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