The left electrode will be positive due to the lower concentration in the left half-cell. The exact voltage depends on the standard cell potential, which is not provided.
The electrode that will be positive in this galvanic cell is the left electrode. The left half-cell has a lower concentration of
, specifically a 350. mM solution, compared to the right half-cell's 3.50 M solution. According to the Nernst equation, as the concentration of the metal ion in the half-cell decreases, the cell potential increases, making the left electrode positive.
The Nernst equation is given by:
![\[ E = E^\circ - \left( (RT)/(nF) \right) \ln \left( ([M^2+])/([M^2+]) \right) \]](https://img.qammunity.org/2024/formulas/chemistry/college/7gmvdyogruhzg5t2iku1apvcm9k94nbeab.png)
Where:
is the cell potential,
is the standard cell potential,
is the ideal gas constant (8.314 J/(mol·K)),
is the temperature in Kelvin,
is the number of moles of electrons transferred in the reaction,
is Faraday's constant (96,485 C/mol),
is the concentration of
in the left half-cell,
is the concentration of
in the right half-cell.
Given the conditions and concentrations, the left electrode will be positive.
As for the voltage displayed on the voltmeter, without specific values for the standard cell potential
, it's challenging to provide an exact voltage. However, the voltmeter would show a positive value, reflecting the higher cell potential due to the lower concentration in the left half-cell. The exact value would require more information, such as the standard cell potential.