A current of 10.0A was passed for 1 hour through a solution of copper (ll) sulphate between the copper electrodes. Which of the following observation would be correct?
(1 Faraday = 96,500 coulombs) (i) 0.186 mole of copper is dissolved from the anode. (ii) 0.372 mole of copper is deposited on the cathode. (iii) the original Cu2+ ion concentration of the solution remains unchanged. (iv) the original Cu2+ ion concentration of the solution decreases.
Cu\(^{2+}\) + 2e- → Cu
1mole of electron = 1F = Number of moles of Cu ions
2moles = 2F = 1 mole of Cu
2 x 96500C = I mole of Cu
10 x 3600C = x moles
X = \(\frac{36000}{193000}\)
X = 0.186 mole of Cu dissolved from the anode.
Also, during electrolysis, copper ions are reduced to copper metals at the cathode(negative electrode). This, however, decreases the concentration of Cu ions present in the solution due to the reduction reaction at the cathode.
These make option C the correct answer.
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}