The following voltaic cells were used in potentiometric analyses, what was the value of E in each case; (give the value involts to 3 decimal places).(a) Ag(s) | Ag (1 M) || Sn (0.086 M), Sn (0.54 M) | Pt(s)Reference electrode potentials (E ):AgCl(s) + e → Ag(s) + Cl (sat’d KCl) E° = 0.197 VHg Cl + 2e → 2Hg + 2Cl (sat’d KCl) E = 0.242 V(= SCE)

I'm sorry, but there seems to be some confusion in your question. The cell you've described is a silver-tin cell, but the reference electrode potentials you've provided are for silver chloride and mercury chloride. Could you please provide the standard reduction potentials for silver and tin?

In general, the cell potential (E) can be calculated using the Nernst equation:

E = E° - (RT/nF) * ln(Q)

where: E° is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred in the redox reaction, F is the Faraday constant, and Q is the reaction quotient.

The standard cell potential (E°) can be calculated from the standard reduction potentials of the two half-reactions. The reaction quotient (Q) can be calculated from the concentrations of the species in the cell.

Once you have these values, you can substitute them into the Nernst equation to calculate the cell potential.

I'm sorry, but the question seems to be incomplete. The cell notation provided does not match the half-reactions given in the reference electrode potentials. Could you please provide the correct half-reactions for the cell notation given? This will help in calculating the cell potential (E) using the Nernst equation.