For the voltaic cell Cr(s)/Cr3+(aq)//Ni+2 (aq)/Ni(s), list the reactions at the cathode, anode, and the overall net reaction.

Answer 1

Hi Tara,
Alright, by convention, we usually write [Oxidation]/[Reduction]in notation. Hence, the oxidation half-cell is the Cr/Cr3+ half-cell while the reduction half-cell is the Ni2+/Ni half-cell. [Alternatively, you can compare E(not) values such that the half-cell with the more negative E(not) value is the oxidation half-cell and the half-cell with the more positive E(not) value is the reduction half-cell.]
We know that for a voltaic cell, the anode is where oxidation (loss of electrons) takes place:
Anode (Oxidation): Cr(s) --> Cr3+(aq) + 3e [Must add 3 electrons to right hand side to balance the charges on both sides of the equation]
We also know that for a voltaic cell, the cathode is where reduction (gain of electrons) takes place:
Cathode (Reduction): Ni2+(aq) + 2e --> Ni(s)[Must add 2 electrons to left hand side to balance the charges on both sides of the equation]
To get the overall equation, balance the number of electrons at the cathode and anode. We note that the lowest common multiple of 2 and 3 is 6.
Hence, to "Cr(s) --> Cr3+(aq) + 3e", we multiply both sides of the equation by 2 to get 2Cr(s) --> 2Cr3+ (aq) + 6e
Similarly, to "Ni2+(aq) + 2e --> Ni(s)", we multiply both sides of the equation by 3 to get 3Ni2+(aq) + 6e --> 3Ni (s)
Then we add up both half equations to get 2Cr(s) + 3Ni2+(aq) + 6e --> 2Cr3+(aq) + 6e + 3Ni (s)
Then cancel out the 6e on both sides to get a final overall net equation: 2Cr(s) + 3Ni2+(aq) --> 2Cr3+(aq) + 3Ni (s)[Note that the final overall net equation cannot have any electrons!]
Hope I helped! (:

-J

Answer 2

Thank you! This helped a lot!