2 Li(s) +Cl₂→ 2 Li⁺ (aq) + 2Cl⁻ (aq)
The cell potential of the reaction above is +4.40V
<em><u>calculation</u></em>
Cell potential =∈° red - ∈° oxidation
in reaction above Li is oxidized from oxidation state 0 to +1 therefore the∈° oxid = -3.04
Cl is reduce from oxidation state 0 to -1 therefore the ∈°red = +1.36 V
cell potential is therefore = +1.36 v -- 3.04 = + 4.40 V
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Explanation:
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Answer:
inerpreting an author's words and using imagination to picture events
Explanation:
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