Calculate the cell potential for the following galvanic cell where a nickel anode is placed in a 1.0M Ni(NO3)2 solution and a silver cathode is placed in a 1.0M AgNO3 solution. The balanced equation for the cell reaction is: Ni(s) + 2 Ag+(aq) → Ni2+(aq) +2 Ag(s) Assume that standard reduction potentials for each half-reaction are as follows: Ni2+(aq) + 2 e- ⇌ Ni(s) E° = -0.25V Ag+(aq) + e- ⇌ Ag(s) E° = +0.80V What is the cell potential at standard conditions (25°C and 1 atm)?

Question

Calculate the cell potential for the following galvanic cell where a nickel anode is placed in a 1.0M Ni(NO3)2 solution and a silver cathode is

Answer 1

To calculate the cell potential at standard conditions, we can use the Nernst equation. However, since the concentrations of the solutions are both 1.0 M, the equation simplifies to:E_cell = E_cathode - E_anodeIn this case, the nickel anode has a standard reduction potential of -0.25 V, and the silver cathode has a standard reduction potential of +0.80 V. Plugging these values into the equation, we get:E_cell =  +0.80 V  -  -0.25 V  = 0.80 V + 0.25 V = 1.05 VSo, the cell potential at standard conditions  25C and 1 atm  for this galvanic cell is 1.05 V.