Based on the half-reactions and their respective standard reduction potentials below, what is the standard cell potential for the reaction that is expected to occur?
Fe3+(aq) + e- → Fe2+(aq) 0.77 V
Sn4+(aq) + 2 e- → Sn2+(aq) 0.15 V
Pb2+(aq) + 2 e- → Pb(s) -0.13 V
Answers:
0.64 V
0.90 V
0.28 V
1.03 V
I don't get the question because there are three possible answers but apparently only one question.
Fe3+(aq) + e- → Fe2+(aq) 0.77 V
Sn2+(aq) ==> Sn4+(aq) + 2 e- - 0.15 V
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add the two and E = 0.77 - 0.15 = + 0.62
One possible reaction but 0.62 is not a possible answer.
Do the same for equations 1 and 3 (turn 3 around) to get E = 0.77 + 0.13 = + 0.9. Reaction is possible and this answer is one of the choices.
Do the same for equations 2 and 3 (turn 3 around) to get E = 0.15 + 0.13 = 0.28. Another possible reaction and this is one of the choices.
Hope this helps you with whatever the question happens to be.
Why did the electron bring a map to the party? Because it was lost and needed directions! Now, let's find the standard cell potential for this reaction, shall we? We simply need to add up the reduction potentials. Using the half-reactions provided, we can see that the Fe3+/Fe2+ half-reaction has a reduction potential of 0.77 V, the Sn4+/Sn2+ half-reaction has a reduction potential of 0.15 V, and the Pb2+/Pb half-reaction has a reduction potential of -0.13 V. Keep in mind that the Pb2+/Pb half-reaction is a reduction, so we'll need to flip the sign of its reduction potential. Now, let's do some math! 0.77 V + 0.15 V + (-0.13 V) equals 0.79 V. Voilà! The standard cell potential for this reaction is 0.79 V. So, the closest answer is 0.90 V. Keep in mind, though, that electrochemistry can be quite shocking sometimes!
To find the standard cell potential for the reaction, we need to subtract the standard reduction potential of the anode half-reaction from the standard reduction potential of the cathode half-reaction.
The anode half-reaction involves the oxidation of Fe2+ to Fe3+, and the reduction potential is 0.77 V (given).
The cathode half-reaction involves the reduction of Sn4+ to Sn2+, and the reduction potential is 0.15 V (given).
Since the anode half-reaction involves the oxidation, the reduction potential should be negated (-0.77 V).
Now, let's subtract the anode potential from the cathode potential:
0.15 V - (-0.77 V) = 0.15 V + 0.77 V = 0.92 V
Therefore, the standard cell potential for the reaction is 0.92 V.
None of the provided answers match this value exactly, but the closest option is 0.90 V.
To find the standard cell potential for the reaction, we need to combine the half-reactions and their respective standard reduction potentials.
The standard cell potential, Ecell, is calculated by taking the difference between the standard reduction potentials of the reduction and oxidation half-reactions.
In this case, the reduction half-reactions are:
Fe3+(aq) + e- → Fe2+(aq) with a standard reduction potential of 0.77 V
Sn4+(aq) + 2e- → Sn2+(aq) with a standard reduction potential of 0.15 V
The oxidation half-reaction is:
Pb2+(aq) + 2e- → Pb(s) with a standard reduction potential of -0.13 V
To balance the equations, we need to multiply the reduction half-reactions by the appropriate coefficients such that the number of electrons gained equals the number of electrons lost in the oxidation half-reaction:
2Fe3+(aq) + 2e- → 2Fe2+(aq) with a standard reduction potential of 0.77 V
Sn4+(aq) + 2e- → Sn2+(aq) with a standard reduction potential of 0.15 V
Pb2+(aq) + 2e- → Pb(s) with a standard reduction potential of -0.13 V
Now, add the balanced half-reactions together:
2Fe3+(aq) + 2e- + Sn2+(aq) + Pb(s) → 2Fe2+(aq) + Sn2+(aq) + Pb(s)
Cancel out the common species on both sides of the equation:
2Fe3+(aq) + Pb(s) → 2Fe2+(aq) + Sn2+(aq)
The standard cell potential, Ecell, can be calculated as the sum of the standard reduction potentials for the reduction half-reactions involved:
Ecell = Ered(cathode) - Ered(anode)
Ecell = (0.77 V + 0.15 V) - (-0.13 V)
Ecell = 0.92 V
Therefore, the correct answer is 0.92 V.