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Question-10023




Question Number 10023 by Tawakalitu ayo mi last updated on 21/Jan/17
Answered by ridwan balatif last updated on 21/Jan/17
6.2  Ni^(2+) +2e^− →Ni_((s)) →reduction           Zn^(2+) +2e^− →Zn_((s)) →oxidation        E^o cell=E_(reduction) ^o −E_(oxidation) ^o                      =−0.23−(−0.76)                     =−0.23+0.76                     =0.53 V
$$\mathrm{6}.\mathrm{2}\:\:\mathrm{Ni}^{\mathrm{2}+} +\mathrm{2e}^{−} \rightarrow\mathrm{Ni}_{\left(\mathrm{s}\right)} \rightarrow\mathrm{reduction} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{Zn}^{\mathrm{2}+} +\mathrm{2e}^{−} \rightarrow\mathrm{Zn}_{\left(\mathrm{s}\right)} \rightarrow\mathrm{oxidation} \\ $$$$\:\:\:\:\:\:\mathrm{E}^{\mathrm{o}} \mathrm{cell}=\mathrm{E}_{\mathrm{reduction}} ^{\mathrm{o}} −\mathrm{E}_{\mathrm{oxidation}} ^{\mathrm{o}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{0}.\mathrm{23}−\left(−\mathrm{0}.\mathrm{76}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{0}.\mathrm{23}+\mathrm{0}.\mathrm{76} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}.\mathrm{53}\:\mathrm{V} \\ $$
Commented by Tawakalitu ayo mi last updated on 21/Jan/17
wow, God bless you sir.
$$\mathrm{wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by ridwan balatif last updated on 21/Jan/17
6.1 reaction will be spontaneous if E^o >0, so          the answer is  (a).spontaneous  (b).spontaneous  (c).spontaneous  (f).spontaneous
$$\mathrm{6}.\mathrm{1}\:\mathrm{reaction}\:\mathrm{will}\:\mathrm{be}\:\mathrm{spontaneous}\:\mathrm{if}\:\mathrm{E}^{\mathrm{o}} >\mathrm{0},\:\mathrm{so} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right).\mathrm{spontaneous} \\ $$$$\left(\mathrm{b}\right).\mathrm{spontaneous} \\ $$$$\left(\mathrm{c}\right).\mathrm{spontaneous} \\ $$$$\left(\mathrm{f}\right).\mathrm{spontaneous} \\ $$
Commented by Tawakalitu ayo mi last updated on 21/Jan/17
i really appreciate sir.
$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

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