Question

Question 4 5 pts Which of the following occurred during the electrolysis of aqueous potassium iodide? copper was plated onto one of the electrodes O the indicator turned pink at one electrode twice as much gas was formed at one electrode that the other brown color disappears at the other electrode O the indicator on one side turned yellow and the other side turned blue gas bubbles were visible only at one electrode a brown color formed at one electrode gas bubbles at both platinum electrodes

Answer 1

Electrolysis of aqueous solution of potassium diode (k g) The probable reactions at cathode (- ve) rightarrow k^+ (a q) + e^- rightarrow k (s) rightarrow E^circ_k^+/K = -2.92 volt & 2 H_2 O + 2 e rightarrow H_2 (g) + 2 OH^- (a q) rightarrow E^circ_H_2 O/OH^- = -0.83 volt We can see that E^circ_k^+/k < < E^circ_H_2 O/OH^- (much smaller) \r\n As the E^circ values (reduction potential) of reaction 1 is much smaller than that of reaction 2 . Therefore, we may conduce that the reaction (2) goes over reaction (1) and hence the reaction at the cathode is - [reaction (2)] 2 H_2 O + 2 e^- rightarrow H_2 (g) + 2 OH^- [At cathode] At cathode OH^- is former (alkali) So, when an indicator [phenolphthalein] is added its color turned in to pink . \r\n Now, the possible Anode (+ ve) reactions are 2 I^(-) (a q) rightarrow I_2 (g) +
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