A galvanic cell is powered by the following redox reaction: 3Cu^2+(aq) + Cr(OH)_

Question

A galvanic cell is powered by the following redox reaction: 3Cu^2+(aq) + Cr(OH)_3(s) + 5 OH^-(aq) rightarrow 3Cu^+(aq) + CrO_4^2-(aq) + 4H_2O(l) Answer the following questions about this cell. If you need any electrochemical data, be sure you get it from the Write a balanced equation for the half-reaction that takes place at the cathode. Write a balanced equation for the half-reaction that takes place at the anode. Calculate the cell voltage under standard conditions. Round your answer to 2 decimal places. E^0 = V

Explanation / Answer

Galvanic cell

3cu2+ (aq) +cr(oh)3(s) +5OH- (aq)----------à3 CU +(aq)+ cro4 2+ (aq) + 4h2o

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Anode

Oxidation:

Cr(OH)3 (s) -> CrO42- (aq)

Oxidation:

2 [ Cr(OH)3(s) + H2O -> CrO42- (aq)+ 5 H+ + 3 e -]

Cu2+(aq) + 2e- Cu(s). (cathode

Cu(s) ------àCu(OH)2(s)+ 2e

3e + CrO4 2 (aq) ------à Cr(OH)3(s)

H2O(l) + Cr(OH)3 (s) ------à CrO42- (aq)

Balance for hydrogen using H+

H2O(l) + Cr(OH)3 (s) àCrO42- (aq) + 5 H+

Add one OH- to BOTH sides for every H+ ion present.

5 OH- + H2O(l) + Cr(OH)3 (s) à CrO42- (aq) + 5 H+ + 5 OH

Combine the OH- and H+ to form water.

5 OH- + H2O(l) + Cr(OH)3 (s) à CrO4 2- (aq) + 5 H2O(l)

Then cancel out the water on the other side.

5 OH- + Cr(OH)3 (s) à CrO4 2- (aq) + 4 H2O(l)

Balance for charge using electrons.

5 OH- + Cr(OH)3 (s) àCrO4 2- (aq) + 4 H2O(l)

left = - 5                             right = - 2

CrO4 2–(aq) + 4H2O(l) + 3e– Cr(OH)3(s) + 5OH– (aq) E o = –0.13 V

Oxidation:

Cr(OH)3 (s) -> CrO42- (aq)

Oxidation:

2 [ Cr(OH)3(s) + H2O -> CrO42- (aq)+ 5 H+ + 3 e -]

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