According to the question, the cell potential is 0.058V of the given galvanic cell.
In an electrochemical cell, the potential difference between two half-cells is measured by the cell potential or Ecell. Because electrons can move easily between one-half cell and the other, there is a potential difference. Because the chemical reaction is a redox reaction, electrons can travel freely between electrodes. Redox reactions happen when one chemical is reduced while another is oxidized. The substance undergoes oxidation, which results in the loss of one or more electrons, making it positively charged. In contrast, when a substance is reduced, it picks up electrons and becomes negatively charged.
According to the question,
By using the Nernst equation,
E(cell) = E(oxidation potential) + E(reduction potential)
E(Cell) = 0.037+0.021
= 0.058 V
Hence, the cell potential is 0.058 V.
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