Calculate the standard electrode potential for the reaction
Cr2+ +3e=Cr(E=-0.74V
Cr3+ +e-=Cr2+(E=-0.41V)
When i calculated i got 0.33V
You're right
To calculate the standard electrode potential for the overall reaction, we need to consider both half-reactions and their corresponding electrode potentials.
Given:
Half-reaction 1: Cr2+ + 3e- = Cr, with E = -0.74V
Half-reaction 2: Cr3+ + e- = Cr2+, with E = -0.41V
To find the overall standard electrode potential (Eo) for the reaction, we can add the half-reactions together and sum their electrode potentials.
Step 1: Reverse the second half-reaction
Cr2+ + e- = Cr3+
Step 2: Multiply the first half-reaction by 2 to balance the number of electrons
2(Cr2+ + 3e-) = 2(Cr)
Step 3: Add the balanced half-reactions together
2(Cr2+ + 3e-) + Cr2+ + e- = 2(Cr) + Cr3+
This simplifies to:
3(Cr2+) + 4e- = 2(Cr) + Cr3+
Step 4: Add up the corresponding electrode potentials
Eo = -0.74V + (-0.41V) = -0.74V - 0.41V = -1.15V
Therefore, the standard electrode potential (Eo) for the overall reaction is -1.15V.
To calculate the standard electrode potential for the given reaction, you need to use the Nernst equation. The Nernst equation relates the standard electrode potential (E°) with the concentration of reactants (or products) and the Faraday constant. The equation is as follows:
E = E° - (0.0592/n) * log(Q)
Where:
- E is the standard electrode potential
- E° is the standard electrode potential at standard conditions (1 M concentration and 298 K)
- n is the number of electrons transferred in the balanced equation
- Q is the reaction quotient
Let's use this equation to calculate the standard electrode potential for the given reaction.
Given balanced half-reactions:
Cr2+ + 3e- → Cr (E = -0.74 V)
Cr3+ + e- → Cr2+ (E = -0.41 V)
We can see that the number of electrons transferred in the first half-reaction is 3, and in the second half-reaction is 1.
To calculate the standard electrode potential for the overall reaction, you need to combine these two half-reactions. The overall reaction is obtained by multiplying the first half-reaction by 2 and the second half-reaction by 3 to balance the electrons.
2Cr3+ + 2e- → 2Cr2+ (E = 2 * -0.41 V)
3Cr2+ + 9e- → 3Cr (E = 3 * -0.74 V)
Adding both reactions, we get:
2Cr3+ + 3Cr2+ → 2Cr + 6Cr2+ (E = 2 * -0.41 V + 3 * -0.74 V)
Simplifying the equation gives:
2Cr3+ + 3Cr2+ → 2Cr + 6Cr2+ (E = -0.82 V + -2.22 V)
Now, let's plug the values into the Nernst equation:
E = E° - (0.0592/n) * log(Q)
Since we are working under standard conditions, the concentration terms in Q for the reaction are taken as 1.
E = -0.82 V + -2.22 V
E = -3.04 V
Therefore, the standard electrode potential for the given reaction is -3.04 V.