Three moles of electrons are passed through three solutions in succession containing `AgNO_(3),CuSO_(4)` and `AuCl_(3)` respectively. The molar ratio of amounts of cations reduced at cathode will be

Three moles of electrons are passed through three solutions in succession containing AgNO_(3),CuSO_(4) and AuCI_(3) respectively. The molar ratio of amounts of cations reduced at cathode will be:-

Three moles of electrons are passed through three solutions in succession containing AgNO_(3),CuSO_(4) and AuCI_(3) respectively. The molar ratio of amounts of cations reduced at cathode will be:-

Three mole of electrons are passed through three solutions in succession containing AgNO_(3), CuSO_(4) and AuCl_(3) respectively. The ratio of amounts of cations reduced at cathode will be :

If three Faradays of electricity is passed through the solutions of AgNO_(3), CuSO_(4) and AuCl_(3), the molar ratio of the cations deposited at the cathodes will be :

Exactly 0.2 mole electrons are passed through two electrolytic cells in series containing CuSO_(4) and ZnSO_(4) respectively. How many grams of each metal will be deposited on the respective cathodes in the two cells ?

If three faradays of electricity are passed through the solutions of AgNO_3, CuSO_4 and AuCl_3 , the molar ratio of the cations deposited at the cathodes will be

1 mol of electrons passes through molten form of AgNo_(3), CuSo_(4) and AICl_(3) when Ag, Cu and Al are deposited. Their molar ratio will be :