1.70 × 10³ seconds
[tex]\text{Co}^{2+}[/tex] + 2 e⁻ → [tex]\text{Co}[/tex]
It takes two moles of electrons to reduce one mole of cobalt (II) ions and deposit one mole of cobalt.
Cobalt has an atomic mass of 58.933 g/mol. 0.500 grams of Co contains [tex]0.500 / 58.933 = 8.484\times 10^{-3} \; \text{mol}[/tex] of Co atoms. It would take [tex]2 \times 8.484 \times 10^{-3} = 0.01697 \; \text{mol}[/tex] of electrons to reduce cobalt (II) ions and produce the [tex]8.484\times 10^{-3} \; \text{mol}[/tex] of cobalt atoms.
Refer to the Faraday's constant, each mole of electrons has a charge of around 96 485 columbs. The 0.01697 mol of electrons will have a charge of [tex]1.637 \times 10^{3} \; \text{C}[/tex]. A current of 0.961 A delivers 0.961 C of charge in one single second. It will take [tex]1.637 \times 10^{3} / 0.961 = 1.70 \times 10^{3} \; \text{s}[/tex] to transfer all these charge and deposit 0.500 g of Co.
