˙Q oʇ uᴉ ʇᴉ pǝƃƃnld I uǝɥ┴ ˙ǝnlɐʌ dsʞ ǝɥʇ ɯoɹɟ +ᄅuZ ɟo uoᴉʇɐɹʇuǝɔuoɔ ǝɥʇ puᴉɟ oʇ ɥɔᴉoʇs pǝsn puɐ Qƃol(u/ᄅ6ϛ0˙0) - oƎ = llǝɔƎ
˙sǝnlɐʌ oƎ oʍʇ ǝɥʇ ƃuᴉppɐ puɐ(ɹǝʇɐʍ ɟo uoᴉʇɐpᴉxo) uoᴉʇɔɐǝɹ ɟlɐɥ ɹǝɥʇo ǝɥʇ ƃuᴉpuᴉɟ ʎq oƎ punoɟ puɐ ʞƃol(u/ᄅ6ϛ0˙0) - o^Ǝ = llǝɔƎ
ʞƃol(u/ᄅ6ϛ0˙0) = llǝɔƎ
˙˙˙pǝʞɹoʍ ǝʌɐɥ ɥɔᴉɥʍ ɟo ǝuou 'spoɥʇǝɯ ƃuᴉʍolloɟ ǝɥʇ ƃuᴉsn pǝᴉɹʇ ǝʌ,I ɹɐɟ oS
(bɐ)-^HOᄅ+ (s)uZ <-> -^ǝᄅ + (s)ᄅ(HO)uZ
˙uoᴉʇɔɐǝɹ-ɟlɐɥ ƃuᴉʍolloɟ ǝɥʇ ɹoɟ llǝɔƎ puᴉℲ
˙ㄣƖ-^0Ɩx8˙Ɩ sᴉ ᄅ(HO)uZ ɟo dsʞ ǝɥ┴
˙ʇɟǝl sʇdɯǝʇʇɐ ᄅ ǝʌɐɥ ʎluo puɐ ɯǝlqoɹd sᴉɥʇ uo ʞɔnʇs ɯ,I
I'm stuck on this problem and only have 2 attempts left.
The Ksp of Zn(OH)2 is 1.8x10^-14.
Find Ecell for the following half-reaction.
Zn(OH)2(s) + 2e^- <-> Zn(s) +2OH^-(aq)
So far I've tried using the following methods, none of which have worked...
Ecell = (0.0592/n)logK
Ecell = E^o - (0.0592/n)logK and found Eo by finding the other half reaction (oxidation of water)and adding the two Eo values.
Ecell = Eo - (0.0592/n)logQ and used stoich to find the concentration of Zn2+ from the Ksp value. Then I plugged it in to Q.
1 answer
1 answer