A chemist designs a galvanic cell that uses these two half-reactions standard re

Question

A chemist designs a galvanic cell that uses these two half-reactions standard reduction potential half-reaction Ered 0.153 V ??? (aq)+4H2O(l)+3e Cr(OH)3(s)+50H (aq) Answer the following questions about this cell Write a balanced equation for the half-reaction that happens at the cathode ei x10 Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written Do you have enough information to calculate the cell voltage under standard conditions? Yes No If you said it was possible to calculate the cell voltage, do so and enter your answerV here. Be sure your answer has the correct number of significant digits

Explanation / Answer

Answer:

(i) Balanced Reduction half reaction takes place at cathode: 3Cu2+ (aq) + 3e- = 3Cu+(aq)

(ii) Balanced Oxidation half reaction takes place at anode:

                                      Cr(OH)3(s) + 5OH-(aq) = CrO42-(aq) + 4H2O(l) + 3e-

(iii)   Oxidation half reaction: Cr(OH)3(s) + 5OH-(aq) = CrO42-(aq) + 4H2O(l) + 3e-

      Reduction half reaction takes place at cathode: 3Cu2+ (aq) + 3e- = 3Cu+(aq)

Therefore the overall balanced reaction will be:

                Cr(OH)3(s) + 5OH-(aq) + 3Cu2+ (aq) = CrO42-(aq) + 4H2O(l) + 3Cu+(aq)

(iv) We can calculate the cell voltage under standard condition.

The voltage of the cell under standard condition will be:

                     E0cell = (E0Right - E0Left ) = (E0Cathode - E0Anode )

                                                  =E0Cu2+/Cu+ - E0Cr(OH)3(s) + 5OH-(aq) /CrO42-(aq) + 4H2O(l) + 3e-

                                                   = + 0.153 - (- 0.13) = + 0.283 volt

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