| Election Name: | 1853 Rhode Island gubernatorial election |
| Type: | Presidential |
| Ongoing: | no |
| Previous Election: | 1852 Rhode Island gubernatorial election |
| Previous Year: | 1852 |
| Next Election: | 1854 Rhode Island gubernatorial election |
| Next Year: | 1854 |
| Election Date: | 6 April 1853 |
| Nominee1: | Philip Allen |
| Party1: | Democratic Party (United States) |
| Popular Vote1: | 10,471 |
| Percentage1: | 54.40% |
| Nominee2: | William W. Hoppin |
| Party2: | Whig Party (United States) |
| Popular Vote2: | 8,228 |
| Percentage2: | 42.75% |
| Nominee3: | Edward Harris |
| Party3: | Free Soil Party |
| Popular Vote3: | 542 |
| Percentage3: | 2.82% |
| Map Size: | 200px |
| Governor | |
| Before Election: | Philip Allen |
| Before Party: | Democratic Party (United States) |
| After Election: | Philip Allen |
| After Party: | Democratic Party (United States) |
The 1853 Rhode Island gubernatorial election was held on 6 April 1853 in order to elect the Governor of Rhode Island. Incumbent Democratic Governor Philip Allen won re-election against Whig nominee William W. Hoppin and Free Soil nominee Edward Harris.[1]
On election day, 6 April 1853, incumbent Democratic Governor Philip Allen won re-election by a margin of 2,243 votes against his foremost opponent William W. Hoppin, thereby keeping democratic control over the office of Governor. Free Soil nominee Edward Harris also was unable to be elected as Governor for the fourth time, after his election losses in the gubernatorial elections of 1849, 1850 and 1851. Allen was sworn in for his third term on 6 May 1853.[2]