What is the standard cell potential of a cell made of theoretical metals Ms/Ma^+2 and Mb/Mb^+2 if the reduction potentials are -0.19V and -0.85 V, respectively?

+0.66V
-0.66V
+1.04 V***
-1.04V

-1.04v

To determine the standard cell potential of the given cell, you need to use the Nernst equation and the reduction potentials of the half-cells involved. The Nernst equation is:

Ecell = E°cell - (0.0592/n)log(Q)

where Ecell is the standard cell potential, E°cell is the standard cell potential at standard conditions, n is the number of electrons transferred in the balanced equation, and Q is the reaction quotient.

In this case, both half-cells involve the transfer of two electrons (as indicated by the superscripts in the cell notation). Therefore, n = 2.

The cell notation for the given cell is:

Ms(s) | Ma^+2(aq) || Mb^+2(aq) | Mb(s)

To calculate the standard cell potential (E°cell), you subtract the reduction potential of the anode from the reduction potential of the cathode (the anode is where oxidation occurs, and the cathode is where reduction occurs).

Given reduction potentials:
- Reduction potential of Ma^+2/Ma: -0.19 V
- Reduction potential of Mb^+2/Mb: -0.85 V

E°cell = E°cathode - E°anode
E°cell = (-0.85 V) - (-0.19 V)
E°cell = -0.85 V + 0.19 V
E°cell = -0.66 V

So, the correct answer is -0.66V.