Chemistry question- half equations
Hello,
Sometimes in exam questions I'm asked for a half equation and proceed as usual... balance 0 with H20... then H+ etc etc...
But sometimes, The markscheme expects only the starting ion instead of the compound..
for example for HgO to Hg, write and equation to show that it accepts electrons. But the equation only had Hg+.
Sometimes in exam questions I'm asked for a half equation and proceed as usual... balance 0 with H20... then H+ etc etc...
But sometimes, The markscheme expects only the starting ion instead of the compound..
for example for HgO to Hg, write and equation to show that it accepts electrons. But the equation only had Hg+.
Reply 1
Original post
by mitostudentHello,
Sometimes in exam questions I'm asked for a half equation and proceed as usual... balance 0 with H20... then H+ etc etc...
But sometimes, The markscheme expects only the starting ion instead of the compound..
for example for HgO to Hg, write and equation to show that it accepts electrons. But the equation only had Hg+.
Sometimes in exam questions I'm asked for a half equation and proceed as usual... balance 0 with H20... then H+ etc etc...
But sometimes, The markscheme expects only the starting ion instead of the compound..
for example for HgO to Hg, write and equation to show that it accepts electrons. But the equation only had Hg+.
I believe these are from old style A level questions - the newer papers would prefer you to write something like
HgO (s) + 2H^+ (aq) + 2e^- —> Hg (l) + H2O (l)
Reply 2
Hello mitostudent!
When mercuric oxide or mercury(II) oxide (HgO) is dissolved in a strong acid, like hydrochloric acid (HCl), the following reaction occurs:
Mercuric chloride (HgCl2) is a strong electrolyte, which means it dissociates into ions when dissolved in water.
So, you can write:
Hg^2+, not HgO, is the initial species in this instance.
The appropriate half-equation in your case would be
As per your statement, the half equation shows the reduction of Hg^2+ to Hg. Indeed, Hg^2+ gains 2 electrons to form Hg.
General notes:
As you can see, there are two types of half-equations. This depends on whether you look at them from the anode (oxidation) or the cathode side (reduction).
Krgds,
Sandro
When mercuric oxide or mercury(II) oxide (HgO) is dissolved in a strong acid, like hydrochloric acid (HCl), the following reaction occurs:
HgO + HCl => HgCl2 + H2O
Mercuric chloride (HgCl2) is a strong electrolyte, which means it dissociates into ions when dissolved in water.
So, you can write:
HgCl2 ==> Hg^2+ + 2Cl^-
Hg^2+, not HgO, is the initial species in this instance.
The appropriate half-equation in your case would be
Hg^2+ + 2e^- ==> Hg
As per your statement, the half equation shows the reduction of Hg^2+ to Hg. Indeed, Hg^2+ gains 2 electrons to form Hg.
General notes:
1.
At the cathode, positively charged ions acquire electrons (reduction)
2.
At the anode, negatively charged ions lose electrons (oxidation)
An example that is not related to what you have written, but only illustrative: Cl2^- ==> Cl2 + 2e^-
As you can see, there are two types of half-equations. This depends on whether you look at them from the anode (oxidation) or the cathode side (reduction).
Krgds,
Sandro
(edited 1 year ago)
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