Write the equations describing the electrode reactions and the net cell reaction for this electrochemical cell containing indium and cadmium:
This is what I have so far:
anode: In(s) --> In^3+ + 3e-
cathode: Cd^2+ + 2e- --> Cd (s)
Need help writing the balanced net reaction!
DrBob222's answer is wrong.
You mean to put Cd/Cd^2+ is -0.403, not 0.403.
Answer is:
3Cd^2+(aq) + 2In(s) --> 3Cd(s) + 2In^3+(aq)
this is right because the short hand notation is:
In(s)|In^3+ (aq)||Cd^2+ (aq)|Cd(s)
And the Anode is always on the left.
~Information valid. College student experience and in the class now.
Ah, an electrochemical cell with indium and cadmium! Let's see if we can balance out those reactions and create a humorous net reaction for you.
We have:
Anode: In(s) --> In^3+ + 3e-
Cathode: Cd^2+ + 2e- --> Cd(s)
To find the net reaction, we need the same number of electrons in both half-reactions. So, let's multiply the cathode reaction by 3!
Anode: In(s) --> In^3+ + 3e-
Cathode (multiplied by 3): 3Cd^2+ + 6e- --> 3Cd(s)
Now, let's put these reactions together and see what happens!
In(s) + 3Cd^2+ --> In^3+ + 3Cd(s)
Voila! Now that's a balanced net reaction! It's like watching a circus act – the indium and cadmium are performing some fancy chemistry acrobatics.
To write the balanced net cell reaction for this electrochemical cell, you need to combine the anode and cathode half-reactions while ensuring that the number of electrons transferred is equal in both reactions. Additionally, the species involved in the half-reactions should be balanced in terms of mass and charge.
Given the half-reactions you've provided:
Anode: In(s) → In^3+ + 3e-
Cathode: Cd^2+ + 2e- → Cd(s)
Step 1: Multiply the anode half-reaction by 2 to balance the number of electrons transferred:
2 In(s) → 2 In^3+ + 6e-
Step 2: Adjust the stoichiometric coefficients to balance the number of electrons transferred. In this case, dividing the anode reaction by 6 and the cathode reaction by 3 will yield the same number of electrons on each side:
Anode: (1/3) In(s) → (1/3) In^3+ + e-
Cathode: Cd^2+ + 2/3e- → Cd(s)
Step 3: Add the two equations together to obtain the net cell reaction:
(1/3) In(s) + Cd^2+ → (1/3) In^3+ + Cd(s)
Thus, the balanced net cell reaction for the electrochemical cell containing indium and cadmium is:
(1/3) In(s) + Cd^2+ → (1/3) In^3+ + Cd(s)
I believe you need to reverse your cell.
1M Cd/Cd^2+ is 0.403 as an oxidation.
1M In/In^3+ is 0.342 as an oxidation.
Therefore, Cd will be the anode and In the cathode.
3Cd + 2In^3+ ==> 3Cd^2+ + 2In