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.
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
Thank you! This helped a lot!
The voltaic cell Cr(s)/Cr3+(aq)//Ni+2(aq)/Ni(s) consists of two half-cells:
1. At the cathode (the reduction half-reaction):
Ni+2(aq) + 2e- -> Ni(s)
2. At the anode (the oxidation half-reaction):
Cr(s) -> Cr3+(aq) + 3e-
By combining these two half-reactions, we can determine the overall net reaction:
Cr(s) + Ni+2(aq) -> Cr3+(aq) + Ni(s)
To determine the reactions at the cathode, anode, and overall net reaction for the given voltaic cell, we first need to understand the cell representation notation.
The cell notation follows the format:
Anode (oxidation) | Salt Bridge || Cathode (reduction)
Given the voltaic cell:
Cr(s) / Cr3+(aq) // Ni2+(aq) / Ni(s)
We can determine the reactions as follows:
1. Anode (Oxidation):
The anode is where oxidation occurs. In this case, the anode is the half-cell containing the Cr(s) and Cr3+(aq) ions.
The oxidation reaction is given by the conversion of Cr(s) to Cr3+(aq).
The half-reaction at the anode is:
Cr(s) -> Cr3+(aq) + 3e-
2. Cathode (Reduction):
The cathode is where reduction occurs. In this case, the cathode is the half-cell containing the Ni2+(aq) and Ni(s) ions.
The reduction reaction is given by the conversion of Ni2+(aq) to Ni(s).
The half-reaction at the cathode is:
Ni2+(aq) + 2e- -> Ni(s)
3. Overall Net Reaction:
To determine the overall net reaction, we need to balance the two half-reactions by multiplying them with appropriate coefficients so that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction.
Multiplying the oxidation half-reaction (1) by 2, we have:
2Cr(s) -> 2Cr3+(aq) + 6e-
Multiplying the reduction half-reaction (2) by 3, we have:
3Ni2+(aq) + 6e- -> 3Ni(s)
Now, we can add both equations to get the overall net reaction:
2Cr(s) + 3Ni2+(aq) -> 2Cr3+(aq) + 3Ni(s)
Thus, the overall net reaction for the given voltaic cell is:
2Cr(s) + 3Ni2+(aq) -> 2Cr3+(aq) + 3Ni(s)