Question

Thank you, put correct number of significant numbers

Answer 1

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
`M(NO_3)^2`, 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:

`\[ E = E^\circ - \left( (RT)/(nF) \right) \ln \left( ([M^2+])/([M^2+]) \right) \]`

Where:

`\(E\)` is the cell potential,

`\(E^\circ\)` is the standard cell potential,

`\(R\)` is the ideal gas constant (8.314 J/(mol·K)),

`\(T\)` is the temperature in Kelvin,

`\(n\)` is the number of moles of electrons transferred in the reaction,

`\(F\)` is Faraday's constant (96,485 C/mol),

`\([M^2+]\)` is the concentration of
`M(NO_3)^2` in the left half-cell,

`\([M^2+]\)` is the concentration of
`M(NO_3)^2` 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
`(\(E^\circ\))`, 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.