See main article: 1836 United States presidential election.
| Election Name: | 1836 United States presidential election in Ohio |
| Country: | Ohio |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1832 United States presidential election in Ohio |
| Previous Year: | 1832 |
| Next Election: | 1840 United States presidential election in Ohio |
| Next Year: | 1840 |
| Election Date: | November 3 – December 7, 1836 |
| Image1: | William Henry Harrison by James Reid Lambdin, 1835 crop.jpg |
| Nominee1: | William Henry Harrison |
| Party1: | Whig Party (United States) |
| Home State1: | Ohio |
| Running Mate1: | Francis Granger |
| Electoral Vote1: | 21 |
| Popular Vote1: | 104,958 |
| Percentage1: | 51.87% |
| Nominee2: | Martin Van Buren |
| Party2: | Democratic Party (United States) |
| Home State2: | New York |
| Running Mate2: | Richard Johnson |
| Electoral Vote2: | 0 |
| Popular Vote2: | 96,238 |
| Percentage2: | 47.56% |
| Map Size: | 200px |
| President | |
| Before Election: | Andrew Jackson |
| Before Party: | Democratic Party (United States) |
| After Election: | Martin Van Buren |
| After Party: | Democratic Party (United States) |
The 1836 United States presidential election in Ohio took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose twenty-one representatives, or electors to the Electoral College, who voted for President and Vice President.
Ohio voted for Whig candidate William Henry Harrison over Democratic candidate Martin Van Buren. Harrison won Ohio by a narrow margin of 4.31%. Ohio was the home state of William Henry Harrison.