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Question Number 10023 by Tawakalitu ayo mi last updated on 21/Jan/17

Answered by ridwan balatif last updated on 21/Jan/17

$$6.2{\mathrm{Ni}}^{2+}+{2\mathrm{e}}^{-}\to {\mathrm{Ni}}_{\left(\mathrm{s}\right)}\to \mathrm{reduction}$$$${\mathrm{Zn}}^{2+}+{2\mathrm{e}}^{-}\to {\mathrm{Zn}}_{\left(\mathrm{s}\right)}\to \mathrm{oxidation}$$$${\mathrm{E}}^{\mathrm{o}}\mathrm{cell}={\mathrm{E}}_{\mathrm{reduction}}^{\mathrm{o}}-{\mathrm{E}}_{\mathrm{oxidation}}^{\mathrm{o}}$$$$=-0.23-(-0.76)$$$$=-0.23+0.76$$$$=0.53\mathrm{V}$$

Commented by Tawakalitu ayo mi last updated on 21/Jan/17

$$\mathrm{wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by ridwan balatif last updated on 21/Jan/17

$$6.1\mathrm{reaction}\mathrm{will}\mathrm{be}\mathrm{spontaneous}\mathrm{if}{\mathrm{E}}^{\mathrm{o}}>0,\mathrm{so}$$$$\mathrm{the}\mathrm{answer}\mathrm{is}$$$$\left(\mathrm{a}\right).\mathrm{spontaneous}$$$$\left(\mathrm{b}\right).\mathrm{spontaneous}$$$$\left(\mathrm{c}\right).\mathrm{spontaneous}$$$$\left(\mathrm{f}\right).\mathrm{spontaneous}$$

Commented by Tawakalitu ayo mi last updated on 21/Jan/17

$$\mathrm{i}\mathrm{really}\mathrm{appreciate}\mathrm{sir}.$$

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