What is the cell potential of an electrochemical cell that has the half-reactions shown below?Fe3+ + e- Fe2+Cu Cu2+ + 2e-Click for a reduction potential chartA.- 0.43 VB.1.2 VC.1.1 VD.0.43 V

To solve this problem, we need to use the Nernst equation and the standard reduction potentials from the reduction potential chart.

The Nernst equation is: Ecell = E°cell - (RT/nF) * ln(Q)

Where: Ecell is the cell potential we want to find, E°cell is the standard cell potential, R is the gas constant, T is the temperature, n is the number of moles of electrons transferred in the balanced redox reaction, F is the Faraday constant, and Q is the reaction quotient.

However, since we are dealing with standard conditions and no concentrations are given, Q is 1 and the ln(Q) term drops out. So, we only need to find E°cell.

E°cell is calculated by subtracting the standard reduction potential of the reaction at the anode from the standard reduction potential of the reaction at the cathode.

From the reduction potential chart, we can find the standard reduction potentials for the two half-reactions:

Fe3+ + e- -> Fe2+ has a standard reduction potential of +0.77 V Cu -> Cu2+ + 2e- has a standard reduction potential of +0.34 V

However, the Cu -> Cu2+ + 2e- reaction is an oxidation reaction (it loses electrons), so we need to reverse the sign of its standard reduction potential when we use it in our calculation.

So, E°cell = E°cathode - E°anode = (+0.77 V) - (-0.34 V) = 1.11 V

So, the cell potential of the electrochemical cell is 1.11 V. The answer is C. 1.1 V.