Find the Emf for the cell

Cr(gr) | Sn4+(aq, .059 M), Sn2+ (aq, .059 M) || Fe3+ (aq, .15 M), Fe2+ (aq, .015 M) | Pt(s) if the standard cell emf is .77 V at the cathode and .15 V at the anode. Answer in Units of V. If would be nice if you could show me step by step.

To find the electromotive force (emf) for the given cell, we can use the Nernst equation, which relates the emf of a cell to the concentration of the species involved. The Nernst equation is given by:

Ecell = E°cell - (RT/nF) * ln(Q)

Where:
- Ecell is the cell emf
- E°cell is the standard cell emf
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- n is the number of electrons transferred in the balanced redox reaction
- F is the Faraday constant (96,485 C/mol)
- ln is the natural logarithm
- Q is the reaction quotient, which is the ratio of the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients.

Let's break down the given half-reactions and determine the values needed to calculate the emf:

1. Cathode (where reduction occurs):
Cr(gr) + 3e- → Cr(s)

Given:
- [Cr(gr)] (unknown concentration)
- [Sn4+(aq)] = 0.059 M
- [Sn2+(aq)] = 0.059 M
- E°cathode = 0.77 V

2. Anode (where oxidation occurs):
Fe2+(aq) → Fe3+(aq) + e-

Given:
- [Fe3+(aq)] = 0.15 M
- [Fe2+(aq)] = 0.015 M
- [Pt(s)] (inert electrode)
- E°anode = 0.15 V

Now, let's calculate the reaction quotient (Q) for the cell:

Q = ([Sn2+(aq)]^2 * [Cr(gr)]) / ([Fe3+(aq)] * [Fe2+(aq)])

Finally, substitute the given values and the calculated Q into the Nernst equation to find the emf (Ecell) of the cell.