| Election Name: | 1894 United States House of Representatives election in Wyoming |
| Country: | Wyoming |
| Type: | presidential |
| Ongoing: | No |
| Previous Election: | 1892 United States House of Representatives election in Wyoming |
| Previous Year: | 1892 |
| Next Election: | 1896 United States House of Representatives election in Wyoming |
| Next Year: | 1896 |
| Nominee1: | Frank Wheeler Mondell |
| Party1: | Republican Party (United States) |
| Popular Vote1: | 10,068 |
| Percentage1: | 52.64% |
| Nominee2: | Henry A. Coffeen |
| Party2: | Democratic Party (United States) |
| Popular Vote2: | 6,152 |
| Percentage2: | 32.17% |
| Nominee3: | Shakespeare E. Sealey |
| Party3: | People's Party (United States) |
| Popular Vote3: | 2,906 |
| Percentage3: | 15.19% |
| Map Size: | 250px |
| U.S. Representative | |
| Before Election: | Henry A. Coffeen |
| Before Party: | Democratic Party (United States) |
| After Election: | Frank Wheeler Mondell |
| After Party: | Republican Party (United States) |
The Wyoming United States House election for 1894 was held on November 6, 1894. Republican Frank Wheeler Mondell defeated Democratic incumbent Henry A. Coffeen and Populist Shakespeare E. Sealey with 52.64% of the vote making Coffeen the second incumbent Representative from Wyoming to lose reelection.